I thought that the objects would move up the slope because of pulley no?Remember: friction is not a fixed force. I call it an "annoying force" in that if an object is trying to move up the incline, friction acts down the incline to oppose your motion. If the object is sliding down the plane, then the friction will act in the opposite direction, up the incline. And remember the friction is a variable force; can keep increasing, as required, but in this case up to mass x 0.251(the maximum friction possible).
But friction, although annoying sometimes, is something we cannot do without. Without friction, we won't be able to walk or drive our car.
depends on the masses, angle of inclination, etc, so the setup of the pulley could cause the objects to move up the slope, or it could cause it them to move down the slope. it this case is seems like the objects will move up the slope, but the question asks for the magnitude of the force, so sign/direction is omitted hereI thought that the objects would move up the slope because of pulley no?
oh ok yh that makes sense, but why is my w/o wrong?depends on the masses, angle of inclination, etc, so the setup of the pulley could cause the objects to move up the slope, or it could cause it them to move down the slope. it this case is seems like the objects will move up the slope, but the question asks for the magnitude of the force, so sign/direction is omitted here
for one thing, you said that the tension in the two ropes is the same, which isn't always true, so there should be 2 different tension forces. tension forces also always point away from the mass, so the tension force connecting m1 to m2 is in the wrong direction. also, the friction should be acting down the slope, which is in the same direction as the force pushing the masses down the slope, so no negative sign for the friction if u use the sign convention ur using. i would prob suggest to just redraw the free body diagram lol, hopefully that will fix ur working out, someone else who knows physics better can prob give a better responseoh ok yh that makes sense, but why is my w/o wrong?
The tension in the force connecting m1 and m2 wpuld be pointing away from the point of attachment right? So wouldnt the arrow from m1 side and arrow from m2 side both be pointing towards the middle of the rope?for one thing, you said that the tension in the two ropes is the same, which isn't always true, so there should be 2 different tension forces. tension forces also always point away from the mass, so the tension force connecting m1 to m2 is in the wrong direction. also, the friction should be acting down the slope, which is in the same direction as the force pushing the masses down the slope, so no negative sign for the friction if u use the sign convention ur using. i would prob suggest to just redraw the free body diagram lol, hopefully that will fix ur working out, someone else who knows physics better can prob give a better response
why would there be no neg sign for friction? i assigned down the slope as negative abd up thr slope as positive, hence why the weight component and friction is neg and the tension is posfor one thing, you said that the tension in the two ropes is the same, which isn't always true, so there should be 2 different tension forces. tension forces also always point away from the mass, so the tension force connecting m1 to m2 is in the wrong direction. also, the friction should be acting down the slope, which is in the same direction as the force pushing the masses down the slope, so no negative sign for the friction if u use the sign convention ur using. i would prob suggest to just redraw the free body diagram lol, hopefully that will fix ur working out, someone else who knows physics better can prob give a better response
yeah but i think u included both the directions in ur calc for m2 (unless if im reading it wrong). in which case they cancel out and u dont account for tension force at all. so when calculating the tension force between m2 and m1 "from m2's perspective" you need to only have the tension force pointing down the slope hereThe tension in the force connecting m1 and m2 wpuld be pointing away from the point of attachment right? So wouldnt the arrow from m1 side and arrow from m2 side both be pointing towards the middle of the rope?
if the tension between m2 and m1 is down the slope, it has the same sign as friction. then the only positive tension is the one between m2 and m3whyv
why would there be no neg sign for friction? i assigned down the slope as negative abd up thr slope as positive, hence why the weight component and friction is neg and the tension is pos
also ive only learnt that tension is equal everywhere in a rope
im kinda confused on what ur tryna say, what i was trying to do first was finding the net force of the entire system and letting that equal (mass*m1+m2+m*)a to find the acceleration. and then i multiplied that acceleration by m2s massyeah but i think u included both the directions in ur calc for m2 (unless if im reading it wrong). in which case they cancel out and u dont account for tension force at all. so when calculating the tension force between m2 and m1 "from m2's perspective" you need to only have the tension force pointing down the slope here
if the tension between m2 and m1 is down the slope, it has the same sign as friction. then the only positive tension is the one between m2 and m3
ohhh okay i see. idk how u would find the net force on the whole system tho, i was more approaching it from focusing on a single mass. @Drongoski will prob be better to ask coz i haven’t done physics in a whileim kinda confused on what ur tryna say, what i was trying to do first was finding the net force of the entire system and letting that equal (mass*m1+m2+m*)a to find the acceleration. and then i multiplied that acceleration by m2s mass
it’s accelerating upwards bc of newtons 3rd law. when the lift is accelerating upwards, the person in the lift is experiencing a force upwards, which means they will push back down on the lift (and hence scales)Also for this q the acceleration is it downwards or upwards? Because when i use either sign conventions (up + down - or down + up -) i ended up getting either -5.194 or 5.914 ms^-2 which means its accelerating downwards however when i searched it up if the apparent weight is the larger than its actual weight then its accelersting upwards View attachment 43005
I keep gettibg an acceleration downwards but my magnitude is rightit’s accelerating upwards bc of newtons 3rd law. when the lift is accelerating upwards, the person in the lift is experiencing a force upwards, which means they will push back down on the lift (and hence scales)
when doing the calculations u might get acceleration downwards (because you are really calculating the reaction force of the person on the lift) but u then have to justify that the force of the lift is causing the person to experience a force upwards, and the person thus pushes down on the lift due to newtons 3rd law increasing the reading on the scales, so the lift must be accelerating upwardsI keep gettibg an acceleration downwards but my magnitude is right
Hmm im still confused i think ill have to watch a vid on how these elevator qs work since it wasnt rlly explicitly covered in tutorwhen doing the calculations u might get acceleration downwards (because you are really calculating the reaction force of the person on the lift) but u then have to justify that the force of the lift is causing the person to experience a force upwards, and the person thus pushes down on the lift due to newtons 3rd law increasing the reading on the scales, so the lift must be accelerating upwards